alternative proof that a finite integral domain is a field

Proof.

Let $R$ be a finite integral domain and $a\\in R$ with $a\eq 0$ . Since $R$ is finite, there exist positive integers $j$ and $k$ with $j<k$ such that $a^{j}=a^{k}$ . Thus, $a^{k}-a^{j}=0$ . Since $j<k$ and $j$ and $k$ are positive integers, $k-j$ is a positive integer. Therefore, $a^{j}(a^{k-j}-1)=0$ . Since $a\eq 0$ and $R$ is an integral domain, $a^{j}\eq 0$ . Thus, $a^{k-j}-1=0$ . Hence, $a^{k-j}=1$ . Since $k-j$ is a positive integer, $k-j-1$ is a nonnegative integer. Thus, $a^{k-j-1}\\in R$ . Note that $a\\cdot a^{k-j-1}=a^{k-j}=1$ . Hence, $a$ has a multiplicative inverse in $R$ . It follows that $R$ is a field.∎

单词	AlternativeProofThatAFiniteIntegralDomainIsAField
释义	alternative proof that a finite integral domain is a field Proof. Let $R$ be a finite integral domain and $a\\in R$ with $a\eq 0$ . Since $R$ is finite, there exist positive integers $j$ and $k$ with $j<k$ such that $a^{j}=a^{k}$ . Thus, $a^{k}-a^{j}=0$ . Since $j<k$ and $j$ and $k$ are positive integers, $k-j$ is a positive integer. Therefore, $a^{j}(a^{k-j}-1)=0$ . Since $a\eq 0$ and $R$ is an integral domain, $a^{j}\eq 0$ . Thus, $a^{k-j}-1=0$ . Hence, $a^{k-j}=1$ . Since $k-j$ is a positive integer, $k-j-1$ is a nonnegative integer. Thus, $a^{k-j-1}\\in R$ . Note that $a\\cdot a^{k-j-1}=a^{k-j}=1$ . Hence, $a$ has a multiplicative inverse in $R$ . It follows that $R$ is a field.∎
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